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# Exam 03 - Version 059 EXAM 3 Fakhreddine(53605 This...

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Version 059 – EXAM 3 – Fakhreddine – (53605) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 12.0 points 100 mL of a 0.40 M solution of HClO is titrated with a solution of 1.6 M NaOH. What is the pH after the addition of 25.0 mL of the NaOH solution? The ionization constant for HClO is 3 . 5 × 10 - 8 . The final volume of the solution is important. 1. 9.12 2. 10.5 correct 3. 9.87 4. 7.46 5. 7.00 6. 10.02 7. 8.92 8. 3.52 Explanation: [HA] = (0 . 4 mol / L)(100 mL) = 40 mmol [OH - ] = (1 . 6 mol / L)(25 mL) = 40 mmol V fin = 125 mL = 0.125 L K w = 10 - 14 [ClO - ] = 0 . 04 mol A - 0 . 125 L = 0 . 32 m / L Analyzing the reaction using molarities, ClO - + H 2 O HClO + OH - ini 0 . 32 0 0 Δ - x x x fin 0 . 32 - x x x For ClO - , K b = K w K a = 1 × 10 - 14 3 . 5 × 10 - 8 = 2 . 85714 × 10 - 7 x 2 0 . 32 - x = 2 . 85714 × 10 - 7 x 2 + 2 . 85714 × 10 - 7 x - 9 . 14286 × 10 - 8 = 0 Applying the quadratic formula, since b 2 - 4 ac = (2 . 85714 × 10 - 7 ) 2 - 4(1)( - 9 . 14286 × 10 - 8 ) = 3 . 65714 × 10 - 7 , x = - 2 . 85714 × 10 - 7 + 3 . 65714 × 10 - 7 2 = 0 . 000302229 Thus [OH - ] = x = 0 . 000302229 , pOH = - log(0 . 000302229) = 3 . 51966 pH = 14 - pOH = 10 . 4803 002 10.0 points The following solutions are mixed in equal volumes. I) 0.1 M HCl and 0.1 M NaCl II) 0.1 M HOCl and 0.1 M NaCl III) 0.1 M HOCl and 0.1 M NaOCl IV) 0.1 M HCl and 0.1 M NaOCl Which will give (a) buffer solution(s)? 1. III only correct 2. II and III only 3. IV only 4. II and IV only 5. I and II only Explanation: A buffer must contain a weak acid/base conjugate pair. OCl - is the conjugate base of the weak acid HOCl. HCl/Cl - is a stong acid conjugate pair. 003 10.0 points Methyl orange is an indicator with a K a of 10 - 4 . Its acid form HIn is red, while its base form In - is yellow. At pH 6.0, the indicator

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Version 059 – EXAM 3 – Fakhreddine – (53605) 2 will be 1. red. 2. colorless. 3. yellow. correct 4. orange. 5. blue. Explanation: 004 10.0 points Mg(OH) 2 would be LEAST soluble in 1. 1 M NaOH. 2. 2 M HClO 4 . 3. pure water. 4. 1 M Ca(OH) 2 . correct 5. 1 M HNO 3 . Explanation: 005 12.0 points A solution is prepared by mixing 250 mL of 1.00 M CH 3 COOH with 500 mL of 1.00 M NaCH 3 COO. What is the pH of this solution? ( K a for CH 3 COOH = 1 . 8 × 10 - 5 ) 1. 5.05 correct 2. 4.44 3. 5.36 4. 4.58 5. 4.74 Explanation: V CH 3 COOH = 250 mL [CH 3 COOH] = 1.00 M V NaCH 3 COO = 500 mL [NaCH 3 COO] = 1.00 M [CH 3 COOH] = 250 mmol 750 mL [CH 3 COO - ] = 500 mmol 750 mL pH = p K a + log [CH 3 COO - ] [CH 3 COOH] = - log(1 . 8 × 10 - 5 ) + log parenleftbigg 500 250 parenrightbigg = 5 . 04576 006 12.0 points The solubility product constant of PbCl 2 is 1 . 7 × 10 - 5 . What is the maximum concentra- tion of Pb 2+ that can be in ocean water that contains 0.0500 M NaCl?
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